JEE Main 2023 (Online) 13th April Evening ShiftQuestion 1
Let for $A=\begin{bmatrix}1&2&3\\\alpha&3&1\\1&1&2\end{bmatrix}$, $|A|=2$. If $|2\operatorname{adj}(2\operatorname{adj}(2A))|=32^{n}$, then $3n+\alpha$ is equal to
Determinant
Solution
$|A|=2$: expanding, $|A|=1(6-1)-2(2\alpha-1)+3(\alpha-3)=-\alpha-2$, so $-\alpha-2=2\Rightarrow\alpha=-4$. For order 3: $|\operatorname{adj}B|=|B|^2$, $|kB|=k^3|B|$. $|2A|=16$; $|\operatorname{adj}(2A)|=2^8$; $|2\operatorname{adj}(2A)|=2^{11}$; $|\operatorname{adj}(\cdot)|=2^{22}$; $|2\operatorname{adj}(\cdot)|=2^{25}=32^5$. So $n=5$ and $3n+\alpha=15-4=11$. Option C.